4.1.1. Convection in a shallow fluid

When a fluid such as water is heated from below (or cooled from above), it develops overturning motions. It may seem obvious that this must occur, because the tendency of the heating (or cooling) is to make the fluid top-heavy.1 Consider the shallow, horizontally infinite fluid shown in Fig. 4.2. Let the heating be applied uniformly at the base; then we may expect the fluid to have a horizontally uniform temperature, so T = T(z) only. This will be top-heavy (warmer and therefore lighter fluid below cold, dense fluid above). But as we have seen, gravitational forces can be balanced by a vertical pressure gradient in

John William Strutt—Lord Rayleigh (1842—1919)—set the study of convection on a firm theoretical basis in his seminal studies in the 1900s. In one of his last articles, published in 1916, he attempted to explain what is now known as Rayleigh-Benard Convection. His work remains the starting point for most of the modern theories of convection.

hydrostatic balance with the density field (Eq. 3-3), in which case any fluid parcel will experience zero net force, even if heavy fluid is above light fluid. Nevertheless, observations and experiments show that convection develops,2 as sketched in the figure. Why? There are two parts to the question:

1. Why do motions develop when the equilibrium state just discussed has no net forces anywhere?

2. Why are the motions horizontally inhomogeneous when the external forcing (the heating) is horizontally uniform?

The answer is that the motions are not directly forced, but (like many types of motion in the atmosphere and ocean) arise from an instability of the fluid in the presence of heating. We therefore begin our discussion of convection by reminding ourselves of the nature of instability.

For any system that possesses some equilibrium state, instability will arise if, in response to a perturbation, the system tends to drive the perturbation further from the equilibrium state. A simple and familiar example is a ball on a curved surface, as in Fig. 4.3.

A ball that is stationary and exactly at a peak (point A, x = xA) is in equilibrium, but of course this state is unstable. If the ball is displaced a small distance Sx from A, it finds itself on a downward slope, and therefore is accelerated further. To be specific, the component of gravitational acceleration along the slope is, for small slope, -g dh/dx. The slope at xA + Sx is, making a Taylor expansion about x = xA (see Appendix A.2.1), dh. dx

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